c++ - 纯 C++ 代码比内联汇编程序快 10 倍。为什么？

Question

这两个代码片段做同样的事情:将两个 float 组相加并将结果存储回其中。

内联汇编器:

void vecAdd_SSE(float* v1, float* v2) { 
    _asm {
        mov esi, v1
        mov edi, v2
        movups xmm0, [esi]
        movups xmm1, [edi]
        addps xmm0, xmm1
        movups [esi], xmm0
        movups [edi], xmm0
    }
}

纯 C++ 代码:

void vecAdd_Std(float* v1, float* v2) {
    v1[0] = v1[0]+ v2[0];
    v1[1] = v1[1]+ v2[1];
    v1[2] = v1[2]+ v2[2];
    v1[3] = v1[3]+ v2[3];

    v2[0] = v1[0];
    v2[1] = v1[1];
    v2[2] = v1[2];
    v2[3] = v1[3];
}

C++ 代码的反汇编(在 Debug模式下进行的反汇编，因为由于某种原因我无法在 Release模式下查看反汇编):

 void vecAdd_Std(float* v1, float* v2) {
 push        ebp  
 mov         ebp,esp  
 sub         esp,0C0h  
 push        ebx  
 push        esi  
 push        edi  
 lea         edi,[ebp-0C0h]  
 mov         ecx,30h  
 mov         eax,0CCCCCCCCh  
 rep stos    dword ptr es:[edi]  

    v1[0] = v1[0]+ v2[0];
 mov         eax,4  
 imul        ecx,eax,0  
 mov         edx,4  
 imul        eax,edx,0  
 mov         edx,dword ptr [v1]  
 mov         esi,dword ptr [v2]  
 movss       xmm0,dword ptr [edx+ecx]  
 addss       xmm0,dword ptr [esi+eax]  
 mov         eax,4  
 imul        ecx,eax,0  
 mov         edx,dword ptr [v1]  
 movss       dword ptr [edx+ecx],xmm0  
    v1[1] = v1[1]+ v2[1];
 mov         eax,4  
 shl         eax,0  
    v1[1] = v1[1]+ v2[1];
 mov         ecx,4  
 shl         ecx,0  
 mov         edx,dword ptr [v1]  
 mov         esi,dword ptr [v2]  
 movss       xmm0,dword ptr [edx+eax]  
 addss       xmm0,dword ptr [esi+ecx]  
 mov         eax,4  
 shl         eax,0  
 mov         ecx,dword ptr [v1]  
 movss       dword ptr [ecx+eax],xmm0  
    v1[2] = v1[2]+ v2[2];
 mov         eax,4  
 shl         eax,1  
 mov         ecx,4  
 shl         ecx,1  
 mov         edx,dword ptr [v1]  
 mov         esi,dword ptr [v2]  
 movss       xmm0,dword ptr [edx+eax]  
 addss       xmm0,dword ptr [esi+ecx]  
 mov         eax,4  
 shl         eax,1  
 mov         ecx,dword ptr [v1]  
 movss       dword ptr [ecx+eax],xmm0  
    v1[3] = v1[3]+ v2[3];
 mov         eax,4  
 imul        ecx,eax,3  
 mov         edx,4  
 imul        eax,edx,3  
 mov         edx,dword ptr [v1]  
 mov         esi,dword ptr [v2]  
 movss       xmm0,dword ptr [edx+ecx]  
 addss       xmm0,dword ptr [esi+eax]  
 mov         eax,4  
 imul        ecx,eax,3  
 mov         edx,dword ptr [v1]  
 movss       dword ptr [edx+ecx],xmm0  

    v2[0] = v1[0];
 mov         eax,4  
 imul        ecx,eax,0  
 mov         edx,4  
 imul        eax,edx,0  
 mov         edx,dword ptr [v2]  
 mov         esi,dword ptr [v1]  
 mov         ecx,dword ptr [esi+ecx]  
 mov         dword ptr [edx+eax],ecx  
    v2[1] = v1[1];
 mov         eax,4  
 shl         eax,0  
 mov         ecx,4  
 shl         ecx,0  
 mov         edx,dword ptr [v2]  
 mov         esi,dword ptr [v1]  
 mov         eax,dword ptr [esi+eax]  
 mov         dword ptr [edx+ecx],eax  
    v2[2] = v1[2];
 mov         eax,4  
 shl         eax,1  
 mov         ecx,4  
 shl         ecx,1  
 mov         edx,dword ptr [v2]  
 mov         esi,dword ptr [v1]  
 mov         eax,dword ptr [esi+eax]  
 mov         dword ptr [edx+ecx],eax  
    v2[3] = v1[3];
 mov         eax,4  
 imul        ecx,eax,3  
 mov         edx,4  
 imul        eax,edx,3  
 mov         edx,dword ptr [v2]  
 mov         esi,dword ptr [v1]  
 mov         ecx,dword ptr [esi+ecx]  
 mov         dword ptr [edx+eax],ecx  

}

现在我对这些函数进行了时间测量，发现内联汇编代码需要大约 10 倍的时间(在 Release模式下)。有人知道为什么吗？

Answer 1

在我的机器上(VS2015 64位模式)，编译器内联vecAdd_Std并产生

00007FF625921C8F  vmovups     xmm1,xmmword ptr [__xmm@4100000040c000004080000040000000 (07FF625929D60h)]  
00007FF625921C97  vmovups     xmm4,xmm1  
00007FF625921C9B  vcvtss2sd   xmm1,xmm1,xmm4  

测试代码

int main() {
    float x[4] = {1.0, 2.0, 3.0, 4.0};
    float y[4] = {1.0, 2.0, 3.0, 4.0};

    vecAdd_Std(x, y);

    std::cout << x[0];
}